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9 golfers playing in threes playing six rounds. Would like all golfers to play in as many combinations so that we all play with each other as equal as possible.also the combination for the group's to equally play in first tee time.second tee time and third tee time so no one golfer plays in one tee too often. Thankyou

Gary Goetz
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  • Do you mean $3$ threesomes in each of $6$ rounds, so $18$ threesomes in all? – saulspatz Jul 07 '19 at 09:02
  • Yes 3 threesomes over 6 rounds – Gary Goetz Jul 07 '19 at 10:58
  • I will give you 9 names to make combinations. – Gary Goetz Jul 07 '19 at 11:06
  • Gary , Jan, Peter, Dave, Bob, Pam, tulio, Chris w , Chris h, thanks – Gary Goetz Jul 07 '19 at 11:08
  • I tried doing this so that every golfer plays with $4$ of the other golfers twice, and with $4$ of them once, but if my computations are correct this is impossible. However, I was assuming that you would not want to have the same set of threesomes twice. It would be possible to have each golfer play with each of the other golfers once in the first four rounds, and then have the threesomes from the first two rounds again. Then each golfer would play with $4$ of the others twice and with $4$ of them once. Would that be admissible? – saulspatz Jul 07 '19 at 11:19
  • Yes please that's fine – Gary Goetz Jul 07 '19 at 11:21
  • Okay, I need to work out the assignments to tees. It may take a while, but I'll have it for you sometime today. – saulspatz Jul 07 '19 at 11:23
  • Thanks so much. Wait to hear – Gary Goetz Jul 07 '19 at 11:29

1 Answers1

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This is not a full answer, but I can't present it in a comment. As I said in my comments, I wrote a program to try to find a schedule where each player plays in three same threesome twice with $4$ golfers, and plays with each of the other $4$ once. Assuming that we never use the same set of three threesomes twice, this turns out to be impossible. If we can repeat threesomes, then we can pair the golfers like this in the first four rounds:

$$ \matrix{012&345&678\\ 048&156&237\\ 036&147&258\\ 057&138&246}$$

You can confirm that each golfer play once with each of the others. Then we can simply reuse the pairings from the first and second rounds in the fifth and sixth rounds.

Now we come to the question of the tees. I tried to arrange it so that everyone teed off twice from each of the three tees, but that's impossible. Eventually, I was driven to looking for an arrangement where each player tees off at least once from each tee, but even that is impossible!

This problem is a lot harder than it looks on the surface.

The best I've been able to do is to arrange that no one tees of from the same tee more than $$ times.

Tee 1: 0 1 2
Tee 2: 3 4 5
Tee 3: 6 7 8

Round 2
Tee 1: 2 3 7
Tee 2: 1 5 6
Tee 3: 0 4 8

Round 3
Tee 1: 2 5 8
Tee 2: 1 4 7
Tee 3: 0 3 6

Round 4
Tee 1: 2 4 6
Tee 2: 1 3 8
Tee 3: 0 5 7

Round 5
Tee 1: 6 7 8
Tee 2: 3 4 5
Tee 3: 0 1 2

Round 6
Tee 1: 1 5 6
Tee 2: 0 4 8
Tee 3: 2 3 7

If this is okay, I can stick the golfers' names in for you. If not, have you any suggestions as to what I might try instead?

EDIT

Round 1
Tee 1: Gary Pete Bob
Tee 2: Jan  Chris w. Pam
Tee 3: Dave Chris h. Tulio

Round 2
Tee 1: Bob Jan  Chris h.
Tee 2: Pete Pam Dave
Tee 3: Gary Chris w. Tulio

Round 3
Tee 1: Bob Pam Tulio
Tee 2: Pete Chris w. Chris h.
Tee 3: Gary Jan  Dave

Round 4
Tee 1: Bob Chris w. Dave
Tee 2: Pete Jan  Tulio
Tee 3: Gary Pam Chris h.

Round 5
Tee 1: Dave Chris h. Tulio
Tee 2: Jan  Chris w. Pam
Tee 3: Gary Pete Bob

Round 6
Tee 1: Pete Pam Dave
Tee 2: Gary Chris w. Tulio
Tee 3: Bob Jan  Chris h.
saulspatz
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  • That's great. I don't think we will get better than that.yes could you please put names in.this is the order Gary Pete Bob Jan Chris w. Pam Dave Chris h. Tulio thanks. – Gary Goetz Jul 07 '19 at 14:23
  • If my answer solved your problem, you should accept it, so the question shows as answered. – saulspatz Jul 07 '19 at 15:22